Stirling formula on $\frac{\gamma(\frac{1}{2}s)}{\gamma(\frac{1-s}{2})} = O(|t|^{\sigma - \frac{1}{2}})$

Question

I tried to prove $\frac{\Gamma(\frac{1}{2}s)}{\Gamma(\frac{1-s}{2})} = O(|t|^{\sigma - \frac{1}{2}})$ using stirling formula when $s =\sigma+it$. However, since stirling formular for gamma function is $\Gamma(z) = \sqrt{\frac{2 \pi}{z}}(\frac{z}{e})^{z} (1+O(\frac{1}{z}))$. I tried to apply Stirling's formula directly but I failed to derive. Could you give any hint on it?

Answer 1

Let's try this using the classical expansion of $\;\ln\Gamma$ (in A&S expression $6.1.41$) : $$\ln\,\Gamma(z)\sim \left(z-\frac 12\right)\ln z-z+\frac 12\ln(2\pi)+\frac 1{12z}+\cdots,\quad \text{for}\;\; z\to\infty,\ |\arg z|<\pi$$ (this is an asymptotic expression and the error doesn't exceed the first term omitted)

Let's apply this formula to the logarithm of your fraction as $\;s\to\infty$ : \begin{align} L(s)&:=\ln\left(\frac{\Gamma\bigl(\frac{s}{2}\bigr)}{\Gamma\bigl(\frac{1-s}{2}\bigr)}\right)\\ &=\ln\,\Gamma\left(\frac{s}{2}\right)-\ln\,\Gamma\left(\frac{1-s}{2}\right)\\ &=\left(\frac {s-1}2\right)\ln \frac s2-\frac s2-\left(-\frac {s}2\right)\ln \frac {1-s}2+\frac {1-s}2+O\left(\frac 1s\right)\\ &=\left(\frac s2\right)\ln\left(\frac s2\frac {1-s}2\right)-\frac 12\ln \frac s2+\frac 12-s+O\left(\frac 1s\right)\\ \end{align} This is not entirely clear in your question but I'll suppose that the real part of $\;s=\sigma+it\,$ i.e. $\sigma$ is bounded while $t$ goes to $+\infty$. We may then expand the first logarithm as : $$\ln\left(\frac {s\,(1-s)}4\right)=\ln\left(\frac {(\sigma+it)(1-\sigma-it)}4\right)=\ln\left(\frac {t^2-it(2\sigma-1)-\sigma\,(\sigma-1)}4\right)=\ln\left(\frac {t^2}4\right)+\ln\left(1-i\frac{2\sigma-1}t-\frac{\sigma\,(\sigma-1)}{t^2}\right)=2\ln \frac t2-i\frac{2\sigma-1}t+O\left(\frac{\sigma^2}{t^2}\right)$$ while the second becomes : $\;\displaystyle \ln \frac s2=\ln \frac {\sigma+it}2=\ln \frac t2+\ln i-i\frac{\sigma}t+O\left(\frac{\sigma^2}{t^2}\right)\,$ so that : \begin{align} L(\sigma+it)&=\left(\frac{\sigma+it}2\right)2\ln \frac t2+\frac{2\sigma-1}2-\frac 12\ln \frac t2-i\frac{\pi}4+\frac 12-\sigma-it+O\left(\frac 1t\right)\\ &=\left(\sigma-\frac 12+it\right)\ln \frac t2-i\left(t+\frac{\pi}4\right)+O\left(\frac 1t\right)\\ \end{align}

The exponential of this last expression will give you : $$\frac{\Gamma\left(\frac{s}{2}\right)}{\Gamma\left(\frac{1-s}{2}\right)}\sim \left(\frac t2\right)^{\large{\sigma-\frac 12+it}}\;e^{\large{-i\left(t+\frac{\pi}4\right)}}$$

Of course taking the absolute value will remove the remaining imaginary terms.